If a, b, c are in A.P. and a, b, d are in G.P., then prove that a, a − b, d − c are in G.P.

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Solution

$a, b and c are in A . P . ∴ 2 b = a + c . . . . . . . (i) Also, a, b and d are in G . P . ∴ b^{2} = a d . . . . . . . (ii) Now, {(a - b)}^{2} = a^{2} - 2 a b + b^{2} = a^{2} - a (a + c) + a d [Using (i) and (ii)] = a d - a c = a (d - c) \Rightarrow {(a - b)}^{2} = a (d - c) Therefore, a, (a - b) and (d - c) are in G . P .$

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If a, b, c are in A.P. and a, b, d are in G.P., then prove that a, a − b, d − c are in G.P.

a, b and c are in A.P.∴ 2b=a+c .......(i)Also, a, b and d are in G.P.∴ b2=ad .......(ii)Now, a-b2=a2-2ab+b2=a2-aa+c+ad Using (i) and (ii)=ad-ac=ad-c⇒a-b2=ad-cTherefore, a, a-b and d-c are in G.P.